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Unformatted text preview: g ( x ) on all of R , we use that it is periodic. Thus, for any x,y R so that | x-y | < (assuming without loss that < 1) we can substract 2 n for some n Z so that x-2 n and y-2 n are in [-2 , 2 ]. Then, | sin x-sin y | = | sin( x-2 n )-sin( y-2 n ) | < since | ( x-2 n )-( y-2 n ) | < . To show nonuniform continuity of fg on R , we shall use that sin x/x 1 as x and Theorem 5.4.2. Indeed, let x n = 2 n and u n = 2 n + 1 n . Then x n-u n 0. x n sin x n = 0 0. But by the sine addition formula, u n sin u n = (2 n + 1 /n ) sin 1 /n . But 2 n sin 1 /n 2 . Thus for n suciently large, | fg ( x n )-fg ( u n ) | = (2 n + 1 /n ) sin 1 /n 1 / 2 . Hence by the Nonuniform Continuity Criterion, fg is not uniformly continous. 1...
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This note was uploaded on 07/15/2010 for the course MATH 521 taught by Professor Metcalfe during the Spring '10 term at University of North Carolina Wilmington.
- Spring '10